Question

1) What kind of acid is the active component in "The Works" toilet bowl cleaner?​

2)If 33 ml of 1.7 M NaOH was required to neutralize 10.0 ml of HCl, what is the molarity of the HCl?

3) What volume (in L) of a 1.16 M NaOH solution is required to neutralize 0.56 mol of HCl?

*Remember to report your answer using the proper significant digits

4) A student has finished the titration of a sample of HBr of unknown concentration with 0.2 M NaOH. The student notices small drops of unreacted NaOH were left on the neck of the flask. This means that these drops were dispensed from the buret but were not mixed with the HBr. How will this affect the calculated molarity of the HBr acid?

Answer 1

1) Hydrochloric acid (HCl) is the active component in "The Works"

2) Use the equation:

V(HCl) * M(HCl) = V (NaOH) * M(NaOH)

M(HCl) = V(NaOH) * M(NaOH)/V(HCl) = 33 * 1.7/10 = 5.6 M

3) # moles of HCl = # moles of NaOH = 0.56 moles

Therefore, V(NaOH) = moles NaOH/ M(NaOH) = 0.56/1.16 = 0.48 L

4) If the drops were dispensed but not mixed with HBr, then it means that the volume of NaOH dispensed would have been more than that actually required. This would then increase the molarity of HBr.